Given the basis of find the matrix

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Homework Statement


Not a homework problem. Typically, we are given a matrix, then asked to find the basis for the kernel or image space of the matrix. I've never seen a problem that did the converse (i.e., given the matrix for the kernel/image space of some matrix, find some matrix).

I was just wondering if anyone's got a good guideline on how to do this.

Say I have the basis for the KERNEL of matrix A formed by the vector:
##\begin{bmatrix}
0\\
1
\end{bmatrix}##.

Homework Equations

The Attempt at a Solution



The # of elements in the vector that forms the basis is 2, so there must be 2 columns in matrix A, but it seems matrix A can 1, 2, 3, 4...and so on # of rows. i.e.,
I can have:
1 0

or
1 0
0 0

or

1 0
0 0
0 0[/B]
.
.
.
and so on.

The basis for the kernel for any of the matrices above would have be spanned by the vector in the OP. Is this the right train of thought?

Also, I don't have a formal procedure on how to find a matrix given, the basis that forms its kernel or image space. I just obtained the previous solution by experience, i guess. Is there a formal way of thinking I should use to obtain the matrix?
 
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pyroknife said:

Homework Statement


Not a homework problem. Typically, we are given a matrix, then asked to find the basis for the kernel or image space of the matrix. I've never seen a problem that did the converse (i.e., given the matrix for the kernel/image space of some matrix, find some matrix).

I was just wondering if anyone's got a good guideline on how to do this.

Say I have the basis for the KERNEL of matrix A formed by the vector:
##\begin{bmatrix}
0\\
1
\end{bmatrix}##.

Homework Equations

The Attempt at a Solution



The # of elements in the vector that forms the basis is 2, so there must be 2 columns in matrix A, but it seems matrix A can 1, 2, 3, 4...and so on # of rows. i.e.,
I can have:
1 0

or
1 0
0 0

or

1 0
0 0
0 0[/B]
.
.
.
and so on.

The basis for the kernel for any of the matrices above would have be spanned by the vector in the OP. Is this the right train of thought?

Also, I don't have a formal procedure on how to find a matrix given, the basis that forms its kernel or image space. I just obtained the previous solution by experience, i guess. Is there a formal way of thinking I should use to obtain the matrix?

You can't do it. As you've said, knowing the kernel doesn't tell you anything about even the dimension of the image space. Knowing the image basis will tell you the dimension. But you certainly can't solve that for a specific matrix. Think about it. A lot of matrices have the same kernel and image.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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